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[附加數]三角(行)數問題(急)

發問:

1.(1+√3)sin^2θ + (1+√3)sinθcosθ+2cos^2θ = 1 2.7sinθcosθ + cos^2θ = 2 3.sinθ - √3cosθ = √2 4.2(sinθ+ cosθ) = √3+1 5.6tan^2θ - 4sin^2θ = 1 6.tan^2θ = 2+ 4cos^2θ 勞煩解釋一下...不太明白如何拆解...

最佳解答:

(1) (1+√3)sin2 θ + (1+√3)sinθcosθ+2cos2 θ = 1 √3sin2 θ + (1+√3)sinθcosθ+2cos2 θ = cos2 θ √3sin2 θ + (1+√3)sinθcosθ+cos2 θ = 0 (√3 sin θ + cos θ) (sin θ + cos θ) = 0 tan θ = -1/√3 or -1 θ = 3π/4, 5π/6, 7π/4 or 11π/6 (2) 7sinθcosθ + cos2 θ = 2 14sinθcosθ + 2cos2 θ = 4 7 sin 2θ + (cos 2θ + 1) = 4 7 sin 2θ + cos 2θ = 3 √50 sin (2θ + 0.142) = 3 sin (2θ + 0.142) = 3/√50 2θ + 0.142 = 0.438, 2.703, 6.721 or 8.987 θ = 0.148, 1.281, 3.290 or 4.422 (3) sinθ - √3cosθ = √2 2 sin (θ - π/3) = √2 sin (θ - π/3) = 1/√2 θ - π/3 = π/4 or 3π/4 θ = 7π/12 or 13π/12 (4) 2(sinθ+ cosθ) = √3+1 4(sinθ+ cosθ)2 = (√3+1)2 4(1 + sin 2θ) = 4 + 2√3 sin 2θ = √3/2 2θ = π/3, 2π/3, 7π/3 or 8π/3 θ = π/6, π/3, 7π/6 or 4π/3 (5) 6 tan2 θ - 4 sin2 θ = 1 6 sin2 θ - 4 sin2 θ cos2 θ = cos2 θ 6 sin2 θ - 4 sin2 θ cos2 θ - cos2 θ = 0 12 sin2 θ - 8 sin2 θ cos2 θ - 2cos2 θ = 0 6 (1 - cos 2θ) - 2 sin2 2θ - (cos 2θ + 1) = 0 5 - 7cos 2θ - 2 (1 - cos2 2θ) = 0 2 cos2 2θ - 7 cos 2θ + 3 = 0 (2 cos 2θ - 1)(cos 2θ - 3) = 0 cos 2θ = 1/2 or 3 (rejected) 2θ = π/3, 5π/3, 7π/3 or 11π/3 θ = π/6, 5π/6, 7π/6 or 11π/6 (6) tan2 θ = 2 + 4 cos2 θ sin2 θ = 2 cos2 θ + 4 (cos2 θ)2 1 - cos2 θ = 2 cos2 θ + 4 (cos2 θ)2 4 (cos2 θ)2 + 3 cos2 θ - 1 = 0 (4cos2 θ - 1)(cos2 θ + 3) = 0 cos2 θ = 1/4 or -3 (rejected) cos θ = 1/2 or -1/2 θ = π/3, 2π/3, 4π/3 or 5π/3 2009-02-14 00:42:35 補充： 全是輔助角 (Aux. angle) 公式轉成的, 通式為: A sin θ + B cos θ = R sin (θ + a) 其中: R = √(A2 + B2) a = cos^-1 (A/R) 或 sin^-1 (B/R)

其他解答:

7 sin 2θ + (cos 2θ + 1) = 4 7 sin 2θ + cos 2θ = 3 √50 sin (2θ + 0.142) = 3 第2題的這3部分不太明白 可解釋一下? 2009-02-12 01:08:36 補充： (3) sinθ - √3cosθ = √2 2 sin (θ - π/3) = √2 以上 這都是 是據什麼轉成的?6FE1C172A843305D

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